Respuesta :
Answer:
a) V = 6.52 10⁻⁵ V, b) Vb = 9.34 10⁻⁵ V
Explanation:
b) When using Gauss's law all the charge is considered concentrated in the inside of the sphere, for which the electrice field is
Q = 3.92 10⁻¹⁵ C
E = k Q / r²
Electric potential and electric field are related
E = - dV / dr
dV = - E dr
Vb-Va = - k Q (1 / rb -1 / ra)
Vb = - 8.99 10⁸ 3.92 10⁻¹⁵ 1 / 0.0377
Vb = 9.34 10⁻⁵ V
a) In this case we must know the electric field for the interior of the sphere, use Gauss's law, the charge inside is
ρ = Q / V
[tex]Q_{int}[/tex] = ρ (4/3 π r³)
Ф = E A = Q_{int} /ε₀
E 4π r² = ρ 4/3 π r³ /ε₀
E = ρ r / 3ε₀
dV = ρ / 3ε₀ ∫ r dr
V = ρ / 3ε₀ r²
We evaluate between the center of the sphere V = 0 and the point R
V = ρ / 3ε₀ ½ (R²)
V = Q 3 / 4π R³ 3ε₀ (R²) = k Q 1 / 2R
V = 8.99 10⁸ 3.92 10⁻¹⁵ 1 / 0.027 2
V = 6.52 10⁻⁵ V
